Let $A$ be an abelian group. We call an element $a\in A$ torsion if there exists a natural number $n$ such that $na = 0$. The set of all torsion elements $T(A)$ of $A$ form a subgroup of $A$, and we can think of $T$ as an endofunctor on the category of abelian groups. Here are some examples:

$T(\Z) = 0$

$T(\Z\oplus \Z/n) = 0\oplus\Z/n\cong\Z/n$

$T(\Q) = 0$

$T(\Q/\Z) = \Q/\Z$

Finding the set of torsion points of an abelian group isn't always easy as in these examples, since abelian groups may not always be written out in such an explicit way. A fascinating and nontrivial result of Barry Mazur is that the torsion subgroup of $E(\Q)$ for an elliptic curve $E$ over $\Q$ is one of fifteen possibilities: $\Z/n$ for $1\leq n\leq 10$ or $n=12$ or $\Z/2\times \Z/2n$ for $1\leq n\leq 4$. Such strange numerology!
Let's call an abelian group torsion if all its elements are torsion. Then, a finite product of torsion groups is again torsion. The infinite product $\prod_{n\geq 2} \Z/n$ has lots of torsion elements, but the element $a(i) = 1$ for all $i$ (the $i$th coordinate is $1$) is not torsion, even though each of its components is torsion. The product $\prod_{n\geq 2} \Z/n$ also provides an example of a group $A$ that is not isomorphic to $T(A)\oplus A/T(A)$, which you might think if you only lived in the finitely-generated world. Indeed, in $\prod_{n\geq 2} \Z/n$ you can have a difference of two torsionfree elements be a nontrivial torsion element! (Can you think of how?)

Abelian groups have an unusual property: $A$ is flat if and only if it $T(A) = 0$. This isn't necessarily true for any old $R$-module for a given ring $R$! For abelian groups, it may be proven by observing that $\mathrm{Tor}_1^\Z(\Q/\Z,A)$ is isomorphic to $T(A)$. So, if $A$ is flat, then $A$ must be torsionfree. What about the converse? If $A$ is torsionfree, then it's flat! Indeed, if $A$ is torsionfree, write it as a direct limit over its finitely generated subgroups: $A = \varinjlim_i A_i$. Since $A$ is an abelian group, $A_i$ is actually free, and hence flat. Since homology commutes with direct limits, we conclude that $\mathrm{Tor}_1(A,B) = 0$ for all abelian groups $B$, so $A$ is flat. Can you figure out how much of this proof generalises to principal ideal domains? What about arbitrary rings?

A simple property of the torsion subgroup $T(A)$ is that the quotient $A/T(A)$ is torsionfree. This is practically self-evident, though one can give the following ridiculous and entertaining proof: first, one checks that the $T(-)$ functor is naturally equivalent to $\mathrm{Tor}_1^\Z(\Q/\Z,-)$. Given that, if we apply the functor $\Q/Z\otimes-$ the short exact sequence $0\to T(A)\to A\to A/T(A)\to 0$ we get the short exact sequence
$$
0\to \mathrm{Tor}_1^\Z(\Q/\Z,T(A))\to \mathrm{Tor}_1^\Z(\Q/\Z,A)\to \mathrm{Tor}_1^\Z(\Q/\Z,A/T(A))\to 0$$
which reduces to $0\to T(A)\to T(A)\to T(A/T(A))\to 0$, with $T(A)\to T(A)$ being an isomorphism, so $T(A/T(A)) = 0$.